LeetCode 1959: Minimum Total Space Wasted With K Resizing Operations Solution

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1959. Minimum Total Space Wasted With K Resizing Operations

Medium

Array Dynamic Programming

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. We will maintain a 2D dp array where dp[i][j] represents the minimum total space wasted up to index i using j resizing operations.

Initialize the dp array with maximum values and set dp[0][0] to 0.
Iterate over the indices of the input array and for each index i:
- Iterate over the number of resizing operations j from 0 to k.
- For each j, calculate the minimum total space wasted by considering resizing at each previous index.
- Update dp[i][j] with the minimum value.
Return the minimum value from dp[nums.length - 1].

Time complexity: O(n^2 * k) where n is the length of nums. Space complexity: O(n * k).

Solution Code

class Solution {
    public int minSpaceWastedKResizing(int[] nums, int k) {
        int n = nums.length;
        int[][] dp = new int[n][k + 1];
        
        for (int i = 0; i < n; i++) {
            Arrays.fill(dp[i], Integer.MAX_VALUE);
        }
        
        dp[0][0] = 0;
        int maxVal = 0;
        for (int i = 0; i < n; i++) {
            maxVal = Math.max(maxVal, nums[i]);
            int sum = 0;
            for (int j = i; j < n; j++) {
                sum += nums[j];
                int wasted = maxVal * (j - i + 1) - sum;
                for (int x = 1; x <= k; x++) {
                    for (int y = i - 1; y < j; y++) {
                        if (y >= 0 && dp[y][x - 1] != Integer.MAX_VALUE) {
                            dp[j][x] = Math.min(dp[j][x], dp[y][x - 1] + wasted);
                        }
                    }
                }
            }
        }
        
        int res = Integer.MAX_VALUE;
        for (int i = 0; i <= k; i++) {
            res = Math.min(res, dp[n - 1][i]);
        }
        
        return res;
    }
}

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